# Change of basis (wrong answer)

Question: Write the change matrix $$Q$$ from the coordinate system $$(x_i',e_i')$$ to the coordinate system $$(x_i'', e_i'')$$, where $$e_1' = (1, 2, 1)^T$$, $$e_2' = (1, 2, 2)^T$$, $$e_3' = (1, 1, 2)^T$$, and $$e_1'' = (1, 0, 1)^T$$, $$e_2'' = (1, 1,0)^T$$, $$e_3'' = (2, 2, 1)^T$$.

My attempt to find the coordinates of $$e_i'$$ with respect to $$\{ e_1'',e_2'',e_3''\}$$

$$e_i' = ae_1''+be_2''+ce_3''$$

$$^t(1, 2, 1)= a ^t(1, 0,1)+ b^t(1, 1, 0)+c^t(2, 2, 1)$$

$$^t(1, 2, 1)= ^t(a+b+2c, b+2c, a+c)$$

Which $$a+b+2c=1$$

$$b+2c=2$$, leading to $$b=2-2c$$

$$a+c=1$$, leading to $$a=1-c$$

So solving it gives be $$a=-1,b=-2,c=2$$

Thus the coordinates of $$e_1'$$ in $$\{ e_1'',e_2'',e_3''\}$$ are $$(-1,-2,2)$$, which this will be the first column of the matrix from $$(x_i',e_i')$$ to $$(x_i'',e_i'')$$

and so on I used

$$e_2' = = a ^t(1, 0,1)+ b^t(1, 1, 0)+c^t(2, 2, 1)$$

$$e_3' = = a ^t(1, 0,1)+ b^t(1, 1, 0)+c^t(2, 2, 1)$$

to find the rest column, which I got matrix of $$Q$$ being

$$\begin{bmatrix}-1&-1&0\\-2&-4&-3\\2&3&2\end{bmatrix}$$

But my colleagues got different answer... is my working wrong? I calculated everything again and again and I do not think it is calculation error but something went wrong instead on the working. Many of my colleagues got

$$\begin{bmatrix}1&2&3\\-2&-2&-3\\2&1&2\end{bmatrix}$$

My colleagues could done it wrong but I doubt so because I'm the only one got this

Was I supposed to do

$$e_i'' = ae_1'+be_2'+ce_3'$$